1. CS4432: Database Systems II
Lecture #13
Professor Elke A. Rundensteiner
CS 4432
query processing

2. Query Processing Query in SQL  Query Plan in Algebra CS 4432

3. Example Data: relation R (A, B, C) relation S (C, D, E) Query:
SELECT B, D
FROM R, S
WHERE R.A = “c” and S.E = 2 and R.C=S.C
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4. SELECT B, D FROM R, S WHERE R.A = “c” and S.E = 2 and R.C=S.C
R A B C S C D E
a x 2
b y 2
c z 2
d x 1
e y 3
Answer B D x
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5. How do we execute query? One idea
- Form Cartesian product of all tables in FROM-clause
- Select tuples that match WHERE-clause
- Project columns that occur in SELECT-clause
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6. R X S R.A R.B R.C S.C S.D S.E
a x 2
a y 2 .
C x 2
Bingo!
Got one...
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7. But ? Performance would be unacceptable!
We need a better approach for:
reasoning about queries,
their execution orders and
their respective costs
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8. Relational Algebra ? Query: SELECT B, D FROM R, S
WHERE R.A = “c” and S.E = 2 and R.C=S.C
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9. Relational Algebra - can be used to describe plans...
Ex: Plan I
B,D
sR.A=“c” S.E=2  R.C=S.C X
R S
OR: B,D [ sR.A=“c” S.E=2  R.C = S.C (RXS)]
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10. Example Instances “Sailors” and “Reserves” relations R1 S1 S2 CS 4432
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11. Relational Algebra Basic operations: Additional operations:
Selection ( ) Selects a subset of rows from relation.
Projection ( ) Deletes unwanted columns from relation.
Cross-product ( ) Allows us to combine two relations.
Set-difference ( ) Tuples in reln. 1, but not in reln. 2.
Union ( ) Tuples in reln. 1 and in reln. 2.
Additional operations:
Intersection, join, division, renaming: Not essential !
Algebra is “closed”: Since each operation returns a relation, operations can be composed !
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12. Projection
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13. Selection
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14. Union, Intersection, Set Difference
Operate on two union-compatible relations:
Same number of fields.
`Corresponding’ fields have same type.
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15. Cross-Product Each row of S1 is paired with each row of R1.
Conflict: Both S1 and R1 have a field called sid.
Renaming operator:
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16. Joins Condition Join : Result schema same as that of cross-product.
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17. Joins Equi-Join: condition contains only equalities.
Result schema only one copy of fields for which equality is specified.
Natural Join: Equijoin on all common fields.
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18. Division Not primitive operator, but useful:
Find sailors who have reserved all boats.
A has 2 fields x and y; B has only field y:
A/B =
i.e., A/B contains all x tuples (sailors) such that for every y tuple (boat) in B, there is an xy tuple in A.
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19. Examples of Division A/B
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20. Expressing A/B Using Basic Operators
Division is useful shorthand.
Idea: For A/B, compute all x values that are not `disqualified’ by some y value in B.
Disqualified x values:
A/B:
all disqualified tuples
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21. Find names of sailors who’ve reserved boat #103
Solution 1:
Solution 2:
Solution 3:
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22. Find names of sailors who’ve reserved a red boat
Information about boat color only available in Boats; so need an extra join:
A more efficient solution:
A query optimizer can find this, given the first solution!
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23. Find sailors who’ve reserved a red or a green boat
Can identify all red or green boats, then find sailors who’ve reserved one of these boats:
Can also define Tempboats using union! (How?)
What happens if is replaced by in this query?
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24. Find sailors who’ve reserved a red and a green boat
Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (sid is a key for Sailors):
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25. Find names of sailors who’ve reserved all boats
Uses division; schemas of input relations to / must be carefully chosen:
To find sailors who’ve reserved all ‘Interlake’ boats:
.....
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26. Relational Algebra representation used to describe plans...
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